It is known that the image of square + (M + 2) x + m of quadratic function y = 2x of X intersects with the axis at two points a and B, and satisfies AB = 4. Find the value of M and the coordinates of point a'b

It is known that the image of square + (M + 2) x + m of quadratic function y = 2x of X intersects with the axis at two points a and B, and satisfies AB = 4. Find the value of M and the coordinates of point a'b

Using the relation between root and coefficient, the square of (x1-x2) = (x1 + x2) square - 4 * x1x2 = 4 square, and the axis X = - b-2a, we can calculate the coordinates
(7y-3z)-(8y-5z)=-y+2z
(7y-3z)-(8y-5z)
=7y-3z-8y+5z
=(7y-8y)-(3z+5z)
=-y-8z
Why + 2Z? I don't think it's wrong,
3z-5z
(7y-3z)-(8y-5z)=-y+2z
Solution: 7y-3z-8y + 5Z + y-2z = 0
2z=0
3Z, 5Z in brackets, it should be 3z-5z, you write it backwards. Pay attention to the symbols in the future.
The third step is wrong. 3Z and 5Z are put in brackets. It should be 3z-5z. You see, in the previous step, 3Z and 5Z are different signs. You put them in brackets and write them the same sign by mistake.
Given the quadratic function y = - 1 / 2x square + 2x-1, what is the area of the triangle ABC when the image intersects the X axis at two points AB and the Y axis at point C
Give me the answer at four tomorrow morning
y=-x²/2+2x-1
Let y = 0 - X & # 178; / 2 + 2x-1 = 0 X & # 178; / 2-2x + 1 = 0 X & # 178; - 4x + 2 = 0 (X-2) & # 178; = 2 X1 = 2 + √ 2 x2 = 2 - √ 2
The intersection of function and x-axis is a (2 + √ 2,0), B (2 - √ 2,0) AB = 2 + √ 2 - (2 - √ 2) = 2 √ 2
Let x = 0, y = - 1, then the intersection coordinates of the function and Y axis are (0, - 1)
The area of triangle ABC is: 1 / 2 × 2 √ 2 ×▏ - 1 ▏ = √ 2
Radical 2
AB is the solution of X when y is 0
0=-1/2x²+2x-1
AB are (2 + root 2,0) (2-root 2,0)
Point C replaces x = 0
Y = - 1
Point C is - 1
The bottom area is 2 * high
The bottom is | 2 + root 2 - (2-root 2) | = 2 root 2
High is | - 1 | = 1
The area is 2
(7y-3z) - (8y-5z) simplification
(7Y-3Z)-(8Y-5Z)
=7Y-3Z-8Y+5Z
=2Z-Y
As shown in the figure, given that the abscissa of the vertex P of the image with quadratic function y = AX2 + BX + C (a ≠ 0) is 4, the X axis of the image intersects point a (m, 0) and point B, and M > 4, then the length of AB is ()
A. 4+mB. mC. 2m-8D. 8-2m
Because the abscissa of the vertex P of the image of the quadratic function y = AX2 + BX + C (a ≠ 0) is 4, the straight line of the parabola symmetry axis is x = 4, intersecting the X axis at point D, so a and B are symmetrical about the symmetry axis. Because point a (m, 0) and M > 4, that is, ad = M-4, so AB = 2ad = 2 (M-4) = 2m-8, so C
Given that x, y and Z are positive integers, and 7x + 2y-5z is a multiple of 11, then 3x + 4Y + 12z divided by 11, the remainder is______ .
Let 7x + 2y-5z = 11m, multiply two sides by 2, and get 14x + 4y-10z = 22m (1) let 3x + 4Y + 12z = n (2) (2) - (1) get - 11x + 22z = n-22m, - 11 (x-2z) = n-22m ∵ the left side is a multiple of 11, ∵ n-22m is also a multiple of 11, ∵ n is also a multiple of 11, ∵ 3x + 4Y + 12z divided by 11, the remainder is 0
It is known that the quadratic function f (x) is an even function with the minimum value of - 1, and the line segment obtained by cutting the x-axis of the image is of length 2
It's better to have a detailed analysis
Let f (x) = AX2 + BX + C, then f (x + 1) = AX2 + (2a + b) x + (a + B + C) f (x + 1) be an even function image with respect to the y-axis symmetry 2A + B = 0. ① the sum of squares of two equations f (x) = 0 is equal to 10. Let two equations be x1, X2, then X1 + x2 = - B / a = 2, x1x2 = C / A, the square of X1 + x2 = 10, C = - 3a
Given that x, y and Z are positive integers, and 7x + 2y-5z is a multiple of 11, then 3x + 4Y + 12z divided by 11, the remainder is______ .
Let 7x + 2y-5z = 11m, multiply two sides by 2, and get 14x + 4y-10z = 22m (1) let 3x + 4Y + 12z = n (2) (2) - (1) get - 11x + 22z = n-22m, - 11 (x-2z) = n-22m ∵ the left side is a multiple of 11, ∵ n-22m is also a multiple of 11, ∵ n is also a multiple of 11, ∵ 3x + 4Y + 12z divided by 11, the remainder is 0
Given the quadratic function y = - X & # 178; + MX-1 and points a (3,0), B (0,3), the necessary and sufficient conditions for finding two different intersections between the image of quadratic function and line AB are obtained
The linear equation of line AB is y = - x + 3, (0
Solving AB's linear equation and substituting it into quadratic function
7x-x=?8y+y=?
6x 9y